1. Design of foundation for Fattouh building in Nablus
Prepared by
Israa Hanani
Noor Sobuh
Nibal Awwad
Supervisor
Dr: Sami Hijjawi

2. Outline 1- Introduction and site description.
2- Load description and type of foundation.
3- Design of foundation.
a. single footing
b. mat foundation
c. pile foundation

3. Introduction and site description
The building foundation or sub structure is that part of a structure which is placed below the surface of the ground and which transmits the superstructure load to the underlying soil ultimately .
It is the part of a structural system that supports and anchors the superstructure of a building .
Building located Fattouh for Aljnied prison and consists of 7 floors. Soil type built by architecture is clay soil . And the allowable bearing capacity is 182 KN/m2.
Based on soil testing, the origin was designed to foundations mat foundation and under it pile .

4. Load description The proposed building has 23 type
of columns based on their dimension .
The distribution of these columns and there corresponding load are shown in table.
As shown from the table below, the columns service loads range from Tons .
Col. no
Dimension
Load (KN) C1
20x60
2110 C4
50x50
2235
C17
30x160
2170
C18
20X100
1540
C19
20x175
2940
C23
20x120
1314

5. Type of foundation
Type
Pile
Mate
Single

6. Design single footing For column 1 (as an example ) :
The dimension of single footing according to applied service load and bearing capacity of soil as follow :
For column 1 (as an example ) :
Area of footing = Pall / Q all
= 2110 / 182 =11.6 m2 (4.5m * 3m)
and repeat it for all column .
From the Fig. we see the overlap between the adjacent footing, so the single footing doesn't fit for this project.

7. Mat Foundation
Area of mat footing = m2

8. 1) We calculate center of loads X p= 1353055.8/ 65563 = 20.64 m
Calculate the center of loading that applied at columns
(col. 1 as an example) :
column Name
Service KN))load Yi Xi
P*Xi
P*Yi C1
2110
11.65
39.9
84189
After that
1) We calculate center of loads
X p= / = m
Yp= / = m
2) Calculate the eccentricity in X and Y directions ex = YP – YG = – = m ey = XP – XG = – 21.3 = m

9. Total soil pressure = 4.53* 13.45 *113.13
q= -(Q/A) – (My X/ Iy ) – (Mxy/Ix) Ix = bh3 / = m4 Iy = m Mx = Q ex = KN.m My = KN.m q = 125 < 182 KN/m2 ……. Ok
We take internal stripe in y- direction with width 4.53m as shown in fig. above
Total soil pressure = 4.53* *113.13
= KN
∑ col loads =
= 6818 KN

10. Shea force diagram

11. B.M.D. Mu = 3615.5 KN.m From the figure of shear and moment we do the
checks and compare the result to the max. value

12. Calculate the thickness of mat footing:-
Width of strip 4.53m
Vu = KN
Assume
D =800 mm h =900 mm
Wide beam shear check :-
Ø vc = (0.75 / 6) (28 ) 0.5 *4530 * 800 / 1000
= KN > Vu ok
Mu = KN.m
ρ = > ρ min ok
As = ρ bd
= * 4530 *800
= mm 2 > As min 1Ø 25 / 200 mm
As min = bh = mm2

13. Check Punching :- Check punching for C 8 :- Pu = 1.4 * 3230 = 4522 KN
ØVc = ( Ø /3 ) (fc )0.5 bod
= ( 0.75 / 3) (28)0.5(2* * 1200 ) (800 ) /1000
= KN > Pu
For columns 11 and 10 we may use drop panel or increase the amount of steel to make it ok for punching

14.   Pile Footing
A pile is a relatively long columnar construction element made of wood, or reinforced concrete, or metal, or a combination thereof. It is embedded into soil to receive and transmit vertical and transmit vertical and inclined loads into the soil or to rock below the ground surface

15. Design Pile Foundation For Column 11
Use All-Pile program
Total Ultimate Capacity (Down)= kN,
Total Ultimate Capacity (Up)= kN
Total Allowable Capacity (Down)= kN,
Total Allowable Capacity (Up)= kN

16. Design Pile Foundation For Column 11
where it's service load equal KN/m2
  Number of Pile
Find the min. length for pile from the equation and satisfied the min.
Ep RI / Su =
So from table min. length = 11* D = 8.8 m
Assume allowable capacity of a 0.8-m-diameter and m-long pile
Total Allowable Capacity (Down)= kN
F=( P/N) +( Mxy/Σyi2) + (Myx/Σxi2) ≤ qall
F= (5788)/N ≤ → N=8→ Use 8 Piles

17. The PLAN show the pile distribution and the spacing

18. Thickness of cap: Pu = 1. 2 DL + 1. 6LL = 7813. 8 KN Ru =Pu/N = 7813
Thickness of cap: Pu = 1.2 DL + 1.6LL = KN Ru =Pu/N = /8 = KN
Wide beam shear :
Vu =976.72*2 = KN ϕVc = (1/6) (√fc) bd /1000 = 0.75(1/6)(√28)(3500)d/1000=
→ d = 850 mm
→ h =1050 mm

19. Punching shear :
For column: Vu= KN ϕVc= 0.75(1/3)(√28)*800*(1350*2+1350*2)/1000= 6071KN > KN → NOT OK SO we increase d= 1000 mm , h=1200 mm ϕVc = 7937 > KN → OK

20. Reinforcement Long direction
Mu = Pu L / 4 = 1172 KN.m , d= 1000 mm, ρ= * 10-3
As = *10-3* 3500 * 1000 = 3122 mm2
As min = bh
= * 3500 * 1200 = 7560 mm For bottom steel > As
→ Use As min
→ use 15 ϕ 25
For Top Steel ->> 7560/2 = 3780 mm2
use 12 ϕ 20
Short direction
= *3500*1200 = 7560 mm2 >As
Use As min → use 15 ϕ 25

21. From Table at Dia. 800 mm Mcapacity = 901.5 KN.m As = 25.1 (10 φ 18)
The figure show the distribution of the steel (a) Show the distribution steel for cap from analysis and (b) Show the distribution of the steel for Pile from classification φ